CUET Preparation Today
The stopping potentials are V1 and V2 with incident lights of wavelength $λ_1$ and $λ_2$ respectively. Then V1 – V2 |
$\frac{hc}{e}(\frac{λ_1λ_2}{λ_1-λ_2})$ $\frac{hc}{e}(\frac{1}{λ_1}-\frac{1}{λ_2})$ $\frac{he}{c}(\frac{1}{λ_1}-\frac{1}{λ_2})$ $\frac{he}{cλ_1-λ_2}(λ_1-λ_2)$ |
$\frac{hc}{e}(\frac{1}{λ_1}-\frac{1}{λ_2})$ |
$eV_1=\frac{hc}{λ_1}-\phi$ $eV_2=\frac{hc}{λ_2}-\phi$ $V_1-V_2=\frac{hc}{e}(\frac{1}{λ_1}-\frac{1}{λ_2})$ |
